4 ms·

You can even go from R to I^d. The problem that I'm having is in getting from R to R^d, which I still don't think is possible - look at the comment where I men

by yarg 3y ago

You can even go from R to I^d. The problem that I'm having is in getting from R to R^d, which I still don't think is possible - look at the comment where I mention "surjections" and "bijections".

Skeime 3y ago
By applying the map from I^d to R^d (using the tan-arctan-construction above in each coordinate) after the map from R to R^d, you get a map from R to R^d.
- yarg 3y ago
  I^n is just a countably infinite subset of R^n.
  - Skeime 3y ago
    No, I^n is also uncountable. In fact, the usual proof for the uncountability of R is really proof of the uncountability of I, and then the natural injection into R is used to extend it to R.